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lesantik [10]
3 years ago
15

What is the pH of a buffer solution that is 0.211 M in hypochlorous acid (HClO) and 0.111 M in sodium hypochlorite? The Ka of hy

pochlorous acid is 3.8 ⋅ 10-8. What is the pH of a buffer solution that is 0.211 M in hypochlorous acid (HClO) and 0.111 M in sodium hypochlorite? The Ka of hypochlorous acid is 3.8 x 10-8.
a. 7.7

b. 0

c. 7.14

d. 14.28

e. 9.05

f. 6.86
Chemistry
1 answer:
arsen [322]3 years ago
7 0

Answer:

c. 7.14

Explanation:

The buffer solution is formed by a weak acid ( hypochlorous acid, HClO) and its conjugate base (hypochlorite ClO⁻, coming from sodium hypochlorite NaClO). We can calculate the pH using the Henderson-Hasselbach equation.

pH = pKa + log [base]/[acid]

pH = -log 3.8 × 10⁻⁸ + log 0.111/0.211

pH = 7.14

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An independent variable is....
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Answer:

B - What we change

Explanation:

Dependent Variable - What we measure

Control Variable - what stays the same

Conclusion - what we conclude

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5 0
3 years ago
The Balmer series, named after Johann Balmer, is a portion of the hydrogen emission spectrum produced from the transitions betwe
horrorfan [7]

Explanation:

The wavelength of the balmer series is calculated using the following steps;

- Find the Principle Quantum Number for the Transition

- Calculate the Term in Brackets

- Multiply by the Rydberg Constant

- Find the Wavelength

The Balmer series in a hydrogen atom relates the possible electron transitions down to the n = 2 position to the wavelength of the emission that scientists observe.

The λ symbol represents the wavelength, and RH is the Rydberg constant for hydrogen, with RH = 1.0968 × 107 m−1

n=7 to n=2

- The principal quantum numbers are 2 and 7.

-  (1/2²) − (1 / n²₂)

For n₂ = 7, you get:

(1/2²) − (1 / n²₂) = (1/2²) − (1 / 7²)

= (1/4) − (1/49)

= 0.2230

- RH = 1.0968 × 107 m−1, to find a value for 1/λ. The formula and the example calculation gives:

1/λ = RH [(1/2²) − (1 / n²₂)]

= 1.0968 × 107 m−1 × 0.2230

= 2445864 m−1

- λ = 1 / 2445864 m−1

= 4.08 × 10−7 m

= 408 nanometers

≈ 410nm

n=6 to n=2

- The principal quantum numbers are 2 and 6.

-  (1/2²) − (1 / n²₂)

For n₂ = 6, you get:

(1/2²) − (1 / n²₂) = (1/2²) − (1 / 6²)

= (1/4) − (1/36)

=  0.2222

- RH = 1.0968 × 107 m−1, to find a value for 1/λ. The formula and the example calculation gives:

1/λ = RH [(1/2²) − (1 / n²₂)]

= 1.0968 × 107 m−1 × 3/16

= 2437090 m−1

- λ = 1 / 2437090 m−1

= 4.10 × 10−7 m

= 410 nanometers

n=5 to n=2

- The principal quantum numbers are 2 and 5.

-  (1/2²) − (1 / n²₂)

For n₂ = 5, you get:

(1/2²) − (1 / n²₂) = (1/2²) − (1 / 5²)

= (1/4) − (1/25)

= 0.21

- RH = 1.0968 × 107 m−1, to find a value for 1/λ. The formula and the example calculation gives:

1/λ = RH [(1/2²) − (1 / n²₂)]

= 1.0968 × 107 m−1 × 0.21

= 2303280 m−1

- λ = 1 / 2303280 m−1

= 4.34 × 10−7 m

= 434 nanometers

n=4 to n=2

- The principal quantum numbers are 2 and 4.

-  (1/2²) − (1 / n²₂)

For n₂ = 4, you get:

(1/2²) − (1 / n²₂) = (1/2²) − (1 / 4²)

= (1/4) − (1/16)

= 0.1875

- RH = 1.0968 × 107 m−1, to find a value for 1/λ. The formula and the example calculation gives:

1/λ = RH [(1/2²) − (1 / n²₂)]

= 1.0968 × 107 m−1 × 0.1875

= 2056500 m−1

- λ = 1 / 2056500 m−1

= 4.86 × 10−7 m

= 486 nanometers

n=3 to n=2

- The principal quantum numbers are 2 and 3.

-  (1/2²) − (1 / n²₂)

For n₂ = 3, you get:

(1/2²) − (1 / n²₂) = (1/2²) − (1 / 3²)

= (1/4) − (1/9)

= 0.13889

- RH = 1.0968 × 107 m−1, to find a value for 1/λ. The formula and the example calculation gives:

1/λ = RH [(1/2²) − (1 / n²₂)]

= 1.0968 × 107 m−1 × 0.13889

= 1523345 m−1

- λ = 1 / 1523345 m−1

= 6.56 × 10−7 m

= 656 nanometers

7 0
2 years ago
Which subatomic part of the atom (a-c) are involved in chemical bonds and reactions?
Marianna [84]
B because it can stabilize
8 0
3 years ago
Carbon dioxide (CO2) is a gas at room temperature and pressure. However, carbon dioxide can be put under pressure to become a "s
Lunna [17]

Answer:

The mass of this 25 mL supercritical CO2 sample has a mass of 11.7g

Explanation:

Step 1: Given data

The supercritical CO2 has a density of 0.469 g/cm³ (or 0.469 g/mL)

The sample hasa volume of 25.0 mL

Step 2: Calculating mass of the sample

The density is the mass per amount of volume

0.469g/cm³ = 0.469g/ml

The mass for a sample of 25.0 mL = 0.469g/mL * 25.0 mL = 11.725g ≈ 11.7g

The mass of this 25 mL supercritical CO2 sample has a mass of 11.7g

5 0
3 years ago
11. What is the mass number
solniwko [45]

Answer:The mass numbr is 22

Explanation:

Mass number=number of protons+ number of neutrons....which is 10+12=22

4 0
3 years ago
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